拉丁方陣
交大應數系
蔡奕正
Definition
A Latin square of order n with entries from an
n-set X is an n＊n array L in which every
cell contains an element of X such that
every row of L is a permutation of X and
every column of L is a permutation of X.
Definition
Let X be a finite set of cardinality n, and let ο be a
binary operation defined on X. We say that the
pair ( X, ο ) is a quasigroup of order n provided
that the following two properties are satisfied:
1.For every x,y  X, the equation x ο z = y has a
unique solution for z  X.
2.For every x,y  X, the equation z ο x = y has a
unique solution for z  X.
Theorem:
Suppose ο is a binary operation defined on a
finite set X of cardinality n. Then ( X, ο ) is
a quasigroup if and only if its operation
table is a Latin square of order n.
Note:
We obtain Latin squares that are the
multiplication tables of groups.
Example:
(c⊕e)⊕d≠c⊕(e ⊕d)
a
b
c
d
e
b
a
e
c
d
c
d
b
e
a
d
e
a
b
c
e
c
d
a
b
Definition:
An orthogonal array OA(n, 3) of order n and
depth 3 is a 3 by n^2 array with the integers 1
to n as entries, such that for any two rows of
the array, the n^2 vertical pairs occurring in
these rows are different.
Definition:
Two Latin squares for which the corresponding
orthogonal arrays have the same three rows
( possibly in different order ) are called
conjugates.
Example:
1111222233334444
1234123412341234
3241142343122134
3241142343122134
1234123412341234
1111222233334444
3
1
4
2
2
4
3
1
4
2
1
3
1
3
2
4
2
4
1
3
4
1
3
2
3
2
4
1
1
3
2
4
1.兩拉丁方陣被稱為等價(equivalent)代表此
兩拉丁方陣可利用換行換列或換元素而
得到相同的拉丁方陣.
2.兩拉丁方陣為等價之例子:
(先作一二列交換,再把元素1及元素3交換)
1
2
3
2
1
3
2
3
1
3
2
1
3
1
2
1
3
2
Note:
 Two orthogonal arrays are called
isomorphic if one can be obtained from the
other by permutations of the elements in
each of the rows and permutations of the
rows and columns of the array. Two Latin
squares for which the corresponding
orthogonal arrays are isomorphic are also
called isomorphic. This means that one is
equivalent to a conjugate of the other.
Example:
1 1 1 2 2 2 3 3 3
1 2 3 1 2 3 1 2 3
1 2 3 2 3 1 3 1 2
1
2
3
2
3
1
3
1
2
1 2 3 2 3 1 3 1 2
1 2 3 1 2 3 1 2 3
1 1 1 2 2 2 3 3 3
1
2
3
3
1
2
2
3
1
2 3 1 3 1 2 1 2 3
3 1 2 3 1 2 3 1 2
1 1 1 2 2 2 3 3 3
2
1
3
3
2
1
1
3
2
Definition:
An n＊n array A with cells which are either
empty or contain exactly one symbol is
called a partial Latin square if no symbol
occurs more than once in any row or
column.
Example:以下兩個partial Latin square都不能
形成完整的拉丁方陣.
1
2
3
1
4
1
1
2
Note:
上述兩個partial Latin square並非真的在實
質上不同， 實際上為conjugates。
1
1
1
2
1
2
3
4
1
2
3
4
1
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4
1
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2
Definition:
If the first k rows of a partial Latin square
( k ≦ n) are filled and the remaining cells are empty,
then A is called a k＊n Latin rectangle.
Theorem: A k＊n Latin rectangle, k＜n, can be
extended to a ( k+1 )＊n Latin rectangle.
Proof:
Let Bj={ a | a不出現在第j行} j=1,2,……,n
∴只須證明B1,B2,……,Bn具SDR即可
W.L.O.G. 取r個集合為B1,B2,……,Br
∵ B1,B2,……,Br共 r(n-k) 個元素(可能重複)
又每個元素至多出現 (n-k) 次
∴ B1,B2,……,Br至少有r個不同元素
∴ B1,B2,……,Bn具SDR
Theorem:L(n) ≧ (n)^2n / n^(n^2)
(We denote by L(n) the total number of different Latin squares of order n)
Proof:
to construct a Latin square of order n, we can take any permutation of 1
to n as the first row. If we have constructed a Latin rectangle with k
rows, then the number of choices for the next row is perB where if
bij=1 if i Bj. By theorem 12.8, this permutation is at least
(n-k)^n．n!/n^n. so we find
L(n) ≧ [n!][ (n-1)^n．n!/n^n][ (n-2)^n．n!/n^n] ……[1^n．n!/n^n]
= (n)^2n / n^(n^2)
Theorem:(due to H. J. Ryser in 1951)
Let A be a partial Latin square of order n in which cell ( i, j )
is filled if and only if i≦r and j≦s. Then A can be
completed to a Latin square if and only if A( i )  r+s-n for
i=1,……,n, where A( i ) denotes the number of elements of
A that are equal to i.
Proof:
()
把完成好的Latin square分成四部份如右
考慮B+D, i在B+D中恰好出現n-s次, i=1,……,n
∴ B( i ) ≦ n-s
又A( i )+ B( i ) = r, ∴ B( i ) = r- A( i )
∴ r- A( i ) ≦ n-s
i.e A( i ) + n-s  r, 即A( i )  r+s-n
A
B
C
D
()
Let B be the (0,1)-matrix of size r＊n with bij=1, if and only if the
element j does not occur in row i of A. Clearly every row of B has sum
n-s. The j-th column of B has sum r-A( j ) ≦ n-s. (∵ A( j )  r+s-n )
By Theorem7.5 ( with d:=n-s ) we have
B=L(s+1)+……+L(n)
where each L(t) is an r＊n (0,1)-matrix with one 1 in each row and at
most one 1 in each column.
Say L(t) = [lij(t)]. Then we fill the cell in position ( i,j ) of A,
i = 1,……,r, j = s+1,……,n, by k if lik(j) = 1.
Thus A is changed into a partial Latin square of order n with r
complete rows, i.e. a Latin rectangle. So this can be completed to a
Latin square of order n.
Theorem:
A partial Latin square of order n with at most n-1 filled cells can
be completed to a Latin square of order n.
(此為有名的Evans conjecture, 且被B. Smetaniuk於1981年證出)
Proof:
case1:
Assume that no row or column contains exactly one filled cell.
Thus the filled cells are contained in at most m rows and columns,
where m:=  n/2 . We may permute rows and columns so that all filled
cells lie in the upper left subarray of order m. ∵ Every m＊m Latin
rectangle can be completed to a Latin square of order n. So it will
suffice to fill in the unfilled cells of the upper left subarray of order m
to get a Latin rectangle. But this is easy, as we have n symbols
available and we may proceed in any order, filled in a cell of the
subsquare with a symbol not yet used in the row or column containing
the cell.
case2:
Assume there is a symbol x of L that occurs only once in L.
詳細證明頗複雜,所以省略
case1之Example:
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1 4
1 3
2 5
3 1 4
1 3 6
2 4 5